每日一练，
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
代码如下：
class Solution {
public:
int level(TreeNode *node)
{
if (node == NULL) return 0;
if (node->left == NULL && node->right == NULL) return 1;
return max(level(node->left),level(node->right)) + 1;
}
void levelPush(TreeNode *node, int height, vector<int> &vec, int cur_height){
if (height == cur_height) vec.push_back(node->val);
if (node->left != NULL) levelPush(node->left, height, vec, cur_height + 1);
if (node->right != NULL) levelPush(node->right, height, vec, cur_height + 1);
}
vector<vector<int> > levelOrder(TreeNode *root) {
int height = level(root);
vector<vector<int> > vvec;
for(int i = 1;i <= height;i++)
{
vector<int> vec;
levelPush(root, i, vec, 1);
vvec.push_back(vec);
}
return vvec;
}
};
